Question

Q. A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base. Prove that the volumes of the parts are in the ratio 1:7:19.​

Answer 1

Answer:

Step-by-step explanation:Since, right triangle ΔABG ~ ΔACF ~ΔADE ( by AA similarity criterion). So corresponding sides are to be proportional.

So, AB/AC = h/2h = 1/2 = BG/CF = r/2r

AB/AD = h/3h = 1/3 = BG/ DE = r/ 3r

Now, we find the volume of each piece.. a smaller cone & 2 frustums

Volume of Cone ABG = 1/3 π r² h ………….(1)

Volume of middle frustum = 1/3 π ( r² + 4r² + 2r² ) h

= 1/3π 7r² h ……………….…….(2)

Volume of next frustum = 1/3 π ( 4r² + 9r² + 6r²) h

= 1/3 π 19r² h …………………….(3)

Now, by finding the ratio of (1),(2)&(3)

we get, (1/3π r² h) : (1/3 π 7r² h) : (1/3 π 19r² h)

= 1:7:19

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